Logarithm Practice Test Questions – Aptitude

log₄​x² + xx + 1 = 2
x(x + 1)x + 1 = 4² = 16
x = 16 (valid, since x + 1 ≠ 0)

log_a 3 = 13 means a^1/3 = 3
Cubing both sides: a = 3³ = 27

Using log A + log B + log C = log(ABC):
= log p²qr × q²pr × r²pq
= log p²q²r²p²q²r² = log 1 = 0

log a - b2 = log √ab
So a - b2 = √ab
a - b = 2√ab
Squaring both sides: (a - b)² = 4ab
a² - 2ab + b² = 4ab
a² + b² = 6ab

log_0.1 0.001 = log(10^-3)log(10^-1) = -3-1 = 3

x = 4^-3/2 = 14^3/2 = 12³ = 18

log₁₀120 = log₁₀(4 × 3 × 10) = log₁₀4 + log₁₀3 + log₁₀10
= 2log₁₀2 + log₁₀3 + 1 = 2x + y + 1

log₁₀60 = log₁₀(2 × 3 × 10) = log₁₀2 + log₁₀3 + log₁₀10 = x + y + 1

0.01 = 10^-2, and 10,000 = 10⁴
log_10^-2(10⁴) = 4 / (-2) = -2

log2[...] = 1 means the inner expression = 2^1 = 2, so log7(x^2-x+37)=2, giving x^2-x+37=49, i.e. x^2-x-12=0, so x=4 or x=-3. x=4 is the listed valid solution.

Expressing everything in terms of ln2 and ln3: log12(27)=a gives ln3/ln2=2a/(3-a). Then log6(16)=4ln2/(ln2+ln3)=4(3-a)/(3+a).

log₃x = 10, so x = 3¹⁰.
log_x y = 100, so y = x¹⁰⁰ = (3¹⁰)¹⁰⁰ = 3¹⁰⁰⁰.

Let log₇ 343 = n.
Then 7ⁿ = 343 = 7³.
So n = 3.

343 = 7 × 7 × 7 = 7³
∴ log₇ 343 = 3 ✓

Let log_0.5 8 = n, so (0.5)ⁿ = 8.
Since 0.5 = 2⁻¹, we get 2⁻ⁿ = 2³.
So −n = 3, that is n = −3.

Base 0.5 = 2⁻¹ → the log changes sign
log₂ 8 = 3 → log_0.5 8 = −3
∴ Answer = −3 ✓

Write 270 = 27 × 10 = 3³ × 10.
log₁₀ 270 = 3 log₁₀ 3 + log₁₀ 10
= 3 × 0.4771 + 1
= 1.4313 + 1 = 2.4313.

270 = 3³ × 10
3(0.4771) + 1 → 1.4313 + 1
∴ log₁₀ 270 = 2.4313 ✓

Write 72 = 8 × 9 = 2³ × 3².
log₁₀ 72 = 3 log₁₀ 2 + 2 log₁₀ 3
= 3 × 0.3010 + 2 × 0.4771
= 0.9030 + 0.9542 = 1.8572.

72 = 2³ × 3²
3(0.3010) + 2(0.4771) → 0.9030 + 0.9542
∴ log₁₀ 72 = 1.8572 ✓

The number of digits in N = (characteristic of log N) + 1.
log (3²⁵) = 25 × log 3 = 25 × 0.4771 = 11.9275.
The characteristic is 11, so the number of digits = 11 + 1 = 12.

Digits = characteristic of log N + 1
25 × 0.4771 = 11.9275 → characteristic 11
∴ Digits = 11 + 1 = 12 ✓

The number of digits in N = (characteristic of log N) + 1.
log (7³⁰) = 30 × log 7 = 30 × 0.8451 = 25.353.
The characteristic is 25, so the number of digits = 25 + 1 = 26.

Digits = characteristic of log N + 1
30 × 0.8451 = 25.353 → characteristic 25
∴ Digits = 25 + 1 = 26 ✓

log₄ x = 3 gives x = 4³ = 64 = 2⁶.
log_x y = 20 gives y = x²⁰.
So y = (2⁶)²⁰ = 2^{6 × 20} = 2¹²⁰.

x = 4³ = 2⁶
y = x²⁰ = (2⁶)²⁰ → powers multiply: 6 × 20 = 120
∴ y = 2¹²⁰ ✓

log₅ x = 4 gives x = 5⁴.
log_x y = 3 gives y = x³.
So y = (5⁴)³ = 5^{4 × 3} = 5¹².

x = 5⁴, y = x³
powers multiply: 4 × 3 = 12
∴ y = 5¹² ✓

Use the change-of-base result 1log_m N = log_N m.
So the sum = log_abc a + log_abc b + log_abc c
= log_abc(a × b × c)
= log_abc(abc) = 1.

1log_m N = log_N m
sum = log_abc a + log_abc b + log_abc c = log_abc(abc)
∴ Sum = 1 ✓

Use 1log_m N = log_N m for every term.
Sum = log₃₆ 2 + log₃₆ 3 + log₃₆ 4 + log₃₆ 9
= log₃₆(2 × 3 × 4 × 9) = log₃₆ 216.
216 = 6³ and 36 = 6², so log₃₆ 216 = 3 log 62 log 6 = 32.

each term: 1log_m 36 = log₃₆ m
sum = log₃₆(2 × 3 × 4 × 9) = log₃₆ 216
216 = 6³, 36 = 6² → 32
∴ Value = 32 ✓

log A + log B + log C = log (A × B × C).
Product = p²qr × q²rp × r²pq = p³q³r³pqr = p²q²r² = (pqr)².
Since pqr = 100, (pqr)² = 100² = 10⁴.
So the value = log 10⁴ = 4.

sum of logs = log of the product
product = (pqr)² = 100² = 10⁴
∴ log 10⁴ = 4 ✓

Sum of logs = log₂ of the product.
Product = a³bc × b³ca × c³ab = a³b³c³a²b²c² = abc = 32.
So the value = log₂ 32 = log₂ 2⁵ = 5.

sum of logs = log₂(product)
product = (abc)³(abc)² = abc = 32
∴ log₂ 32 = 5 ✓

Take log (base 10) of both sides: x log 254 = log 7.
So x = log 7log 25 − log 4.
Now log 25 = 2 log 5 and log 4 = 2 log 2.
Hence x = log 72 log 5 − 2 log 2.

254^x = 7 → x = log 7log 25 − log 4
log 25 − log 4 = 2 log 5 − 2 log 2
∴ x = log 72 log 5 − 2 log 2 ✓

Take log of both sides: x log 8 = log 20, so x = log 20log 8.
log 8 = 3 log 2 = 3 × 0.3010 = 0.9030.
log 20 = log 2 + log 10 = 0.3010 + 1 = 1.3010.
x = 1.30100.9030 = 1.44 (approximately).

x = log 20log 8
log 20 = 1.3010, log 8 = 3(0.3010) = 0.9030
1.3010 ÷ 0.9030 ≈ 1.44
∴ x ≈ 1.44 ✓

If log A = log B (with A and B positive), then A = B.
3x² + 16 = (x − 4)² = x² − 8x + 16
2x² + 8x = 0
2x(x + 4) = 0, so x = 0 or x = −4.
Check: for x = 0 both sides are log 16, and for x = −4 both sides are log 64 (3 × 16 + 16 = 64 = (−8)²). Both values are valid.

log A = log B → A = B
3x² + 16 = x² − 8x + 16 → 2x² + 8x = 0
2x(x + 4) = 0
∴ x = 0, −4 ✓

If log A = log B (with A and B positive), then A = B.
4x² + 9 = (2x − 3)² = 4x² − 12x + 9
The 4x² and 9 cancel on both sides, leaving 12x = 0, so x = 0.
Check: at x = 0 both sides equal log 9, and (2x − 3)² = 9 is positive. So x = 0 is the only solution.

4x² + 9 = 4x² − 12x + 9
cancel 4x² and 9 → 12x = 0
∴ x = 0 ✓

For base 10, 10^{log m} = m, so 10^{log 2} = 2.
log₃(x² − 7) = 2, so x² − 7 = 3² = 9.
x² = 16, so x = 4 or x = −4.
Check: x² − 7 = 9 is positive for both values, so both are valid.

10^{log 2} = 2
log₃(x² − 7) = 2 → x² − 7 = 9 → x² = 16
∴ x = 4 and −4 ✓

For base 10, 10^{log m} = m (valid only when m > 0), so the right side is 3 − x and we need x < 3.
log₂(9 − 2^x) = 3 − x gives 9 − 2^x = 2^{3 − x} = 82^x.
Let t = 2^x: 9 − t = 8t gives t² − 9t + 8 = 0, so (t − 1)(t − 8) = 0 and t = 1 or t = 8.
t = 1 gives x = 0, and t = 8 gives x = 3.
x = 3 is rejected because 3 − x = 0 and log 0 is not defined.
Check x = 0: left side = log₂ 8 = 3 and right side = 10^{log 3} = 3. So x = 0 only.

10^{log(3 − x)} = 3 − x → need x < 3
9 − 2^x = 2^{3 − x}; t = 2^x → t² − 9t + 8 = 0 → t = 1, 8 → x = 0, 3
x = 3 makes 3 − x = 0 (log undefined) → reject
∴ x = 0 only ✓